System of two Equations: $\sqrt x+y=11$ and $\sqrt y+x=7$ A friend of Mine gave me a system of two equations and asked me to solve them $\rightarrow$
$$\sqrt{x}+y=11~~ ...1$$
$$\sqrt{y}+x=7~~ ...2$$
I tried to solve them manually and got this horrendously complicated fourth degree equation $\rightarrow$
$$\begin{align*}
y &= (7-x)^2 ~...\mbox{(from 2)} \\
y &= 49 - 14 x + x^2 \\
\implies 11&= \sqrt{x}+ 49 - 14 x + x^2 ...(\mbox{from 1)}\\
\implies~~ 0&=x^4-28x^3+272x^2-1065x+1444
\end{align*}$$ 
Solving this wasn't exactly my piece of cake but I could tell that one of Solutions would have been 9 and 4 
But my friend kept asking for a formal solution.
I tried plotting the equations and here's what I got $\rightarrow$

So the equations had two pairs of solutions (real ones).
Maybe, Just maybe I think these could be solved using approximations.
So How do i solve them using a formal method (Calculus,Algebra,Real Analysis...) 
P.S.  I'm In high-school.
 A: Assume $x$ and $y$ are integers. Notice that, in this case, if $\sqrt x +y=11$, an integer, then $\sqrt x $ must be an integer. A similar argument can be made for $y$. So if they're integers then they're both perfect squares. Rephrasing in terms of the square roots (still integers) $X=\sqrt x,Y=\sqrt y$
$$X+Y^2=11$$
$$Y+X^2=7$$
subtracting the second equation from the first:
$$X-Y+Y^2-X^2=4$$
$$(X-Y)+(Y-X)(Y+X)=4$$
$$(Y-X)(X+Y-1)=4$$
Both of the brackets are integers, so the only values they can take are the factors of $4$. So either
$$Y-X=2,X+Y-1=2$$
or
$$Y-X=4,X+Y-1=1$$
or$$Y-X=1,X+Y-1=4$$
Solving each of these is simple. The only one that gives positive integer values (the conditions of our little set up here) is the $3^{rd}$ one, which gives the answer you found. Keep in mind that there's nothing wrong with guessing and playing around with the problem first, then coming to a more structured argument later. If you want a full analytic solution you could use the quartic equation on the one you have and rule out the other solutions as involving the wrong branches of $\sqrt x$, but it's pointlessly messy.
A: Once you guessed the solutions, you can easily prove that there are no others. Rewrite the equations as $y=11-\sqrt x=F(x)$ and $x=7-\sqrt y=G(y)$. Note that both $x,y\le 11$, so their square roots are at most $4$, which means that $x,y\ge 3$. Now just observe that $z\mapsto \sqrt z$ is a contraction on $[3,\infty)$ (the difference of values is less than the difference of arguments). Thus, $F$ and $G$ are also contractions whence if we had two different solutions $(x_1,y_1)$ and $(x_2,y_2)$, we would get 
$$
|x_1-x_2|=|G(y_1)-G(y_2)|<|y_1-y_2|=|F(x_1)-F(x_2)|<|x_1-x_2|
$$
which is absurd.
